3.13.52 \(\int \frac {50 x-50 x^2+e^{4 x+x^2} (-200 x^2-100 x^3)+(x-x^3+e^{4 x+x^2} (1-x^2)+e^3 (-1+x^2)+(-1+x^2) \log (x)) \log ^3(-e^3+e^{4 x+x^2}+x-\log (x))}{(e^3 x^2-e^{4 x+x^2} x^2-x^3+x^2 \log (x)) \log ^3(-e^3+e^{4 x+x^2}+x-\log (x))} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{x}+x-\frac {25}{\log ^2\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \]

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Rubi [F]  time = 7.99, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {50 x-50 x^2+e^{4 x+x^2} \left (-200 x^2-100 x^3\right )+\left (x-x^3+e^{4 x+x^2} \left (1-x^2\right )+e^3 \left (-1+x^2\right )+\left (-1+x^2\right ) \log (x)\right ) \log ^3\left (-e^3+e^{4 x+x^2}+x-\log (x)\right )}{\left (e^3 x^2-e^{4 x+x^2} x^2-x^3+x^2 \log (x)\right ) \log ^3\left (-e^3+e^{4 x+x^2}+x-\log (x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(50*x - 50*x^2 + E^(4*x + x^2)*(-200*x^2 - 100*x^3) + (x - x^3 + E^(4*x + x^2)*(1 - x^2) + E^3*(-1 + x^2)
+ (-1 + x^2)*Log[x])*Log[-E^3 + E^(4*x + x^2) + x - Log[x]]^3)/((E^3*x^2 - E^(4*x + x^2)*x^2 - x^3 + x^2*Log[x
])*Log[-E^3 + E^(4*x + x^2) + x - Log[x]]^3),x]

[Out]

x^(-1) + x + 200*Defer[Int][Log[-E^3 + E^(x*(4 + x)) + x - Log[x]]^(-3), x] + 100*Defer[Int][x/Log[-E^3 + E^(x
*(4 + x)) + x - Log[x]]^3, x] - 50*Defer[Int][1/(x*(-E^3 + E^(x*(4 + x)) + x - Log[x])*Log[-E^3 + E^(x*(4 + x)
) + x - Log[x]]^3), x] - 100*Defer[Int][x^2/((-E^3 + E^(x*(4 + x)) + x - Log[x])*Log[-E^3 + E^(x*(4 + x)) + x
- Log[x]]^3), x] + 100*Defer[Int][(x*Log[x])/((-E^3 + E^(x*(4 + x)) + x - Log[x])*Log[-E^3 + E^(x*(4 + x)) + x
- Log[x]]^3), x] - 50*(1 + 4*E^3)*Defer[Int][1/((E^3 - E^(x*(4 + x)) - x + Log[x])*Log[-E^3 + E^(x*(4 + x)) +
x - Log[x]]^3), x] + 100*(2 - E^3)*Defer[Int][x/((E^3 - E^(x*(4 + x)) - x + Log[x])*Log[-E^3 + E^(x*(4 + x))
+ x - Log[x]]^3), x] - 200*Defer[Int][Log[x]/((E^3 - E^(x*(4 + x)) - x + Log[x])*Log[-E^3 + E^(x*(4 + x)) + x
- Log[x]]^3), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 x-50 x^2-100 e^{x (4+x)} x^2 (2+x)-\left (-1+x^2\right ) \left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}{x^2 \left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx\\ &=\int \left (\frac {50 \left (1-\left (1+4 e^3\right ) x+4 \left (1-\frac {e^3}{2}\right ) x^2+2 x^3-4 x \log (x)-2 x^2 \log (x)\right )}{x \left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}+\frac {200 x^2+100 x^3-\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )+x^2 \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}{x^2 \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}\right ) \, dx\\ &=50 \int \frac {1-\left (1+4 e^3\right ) x+4 \left (1-\frac {e^3}{2}\right ) x^2+2 x^3-4 x \log (x)-2 x^2 \log (x)}{x \left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+\int \frac {200 x^2+100 x^3-\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )+x^2 \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}{x^2 \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx\\ &=50 \int \left (-\frac {1}{x \left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}-\frac {2 x^2}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}+\frac {2 x \log (x)}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}-\frac {1+4 e^3}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}-\frac {2 \left (-2+e^3\right ) x}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}-\frac {4 \log (x)}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}\right ) \, dx+\int \left (1-\frac {1}{x^2}+\frac {100 (2+x)}{\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}\right ) \, dx\\ &=\frac {1}{x}+x-50 \int \frac {1}{x \left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+100 \int \frac {2+x}{\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-100 \int \frac {x^2}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+100 \int \frac {x \log (x)}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-200 \int \frac {\log (x)}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+\left (100 \left (2-e^3\right )\right ) \int \frac {x}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-\left (50 \left (1+4 e^3\right )\right ) \int \frac {1}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx\\ &=\frac {1}{x}+x-50 \int \frac {1}{x \left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+100 \int \left (\frac {2}{\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}+\frac {x}{\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )}\right ) \, dx-100 \int \frac {x^2}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+100 \int \frac {x \log (x)}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-200 \int \frac {\log (x)}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+\left (100 \left (2-e^3\right )\right ) \int \frac {x}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-\left (50 \left (1+4 e^3\right )\right ) \int \frac {1}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx\\ &=\frac {1}{x}+x-50 \int \frac {1}{x \left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+100 \int \frac {x}{\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-100 \int \frac {x^2}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+100 \int \frac {x \log (x)}{\left (-e^3+e^{x (4+x)}+x-\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+200 \int \frac {1}{\log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-200 \int \frac {\log (x)}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx+\left (100 \left (2-e^3\right )\right ) \int \frac {x}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx-\left (50 \left (1+4 e^3\right )\right ) \int \frac {1}{\left (e^3-e^{x (4+x)}-x+\log (x)\right ) \log ^3\left (-e^3+e^{x (4+x)}+x-\log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 30, normalized size = 1.07 \begin {gather*} \frac {1}{x}+x-\frac {25}{\log ^2\left (-e^3+e^{4 x+x^2}+x-\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(50*x - 50*x^2 + E^(4*x + x^2)*(-200*x^2 - 100*x^3) + (x - x^3 + E^(4*x + x^2)*(1 - x^2) + E^3*(-1 +
x^2) + (-1 + x^2)*Log[x])*Log[-E^3 + E^(4*x + x^2) + x - Log[x]]^3)/((E^3*x^2 - E^(4*x + x^2)*x^2 - x^3 + x^2
*Log[x])*Log[-E^3 + E^(4*x + x^2) + x - Log[x]]^3),x]

[Out]

x^(-1) + x - 25/Log[-E^3 + E^(4*x + x^2) + x - Log[x]]^2

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fricas [B]  time = 0.76, size = 56, normalized size = 2.00 \begin {gather*} \frac {{\left (x^{2} + 1\right )} \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2} - 25 \, x}{x \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-1)*log(x)+(-x^2+1)*exp(x^2+4*x)+(x^2-1)*exp(3)-x^3+x)*log(-log(x)+exp(x^2+4*x)+x-exp(3))^3+(-
100*x^3-200*x^2)*exp(x^2+4*x)-50*x^2+50*x)/(x^2*log(x)-x^2*exp(x^2+4*x)+x^2*exp(3)-x^3)/log(-log(x)+exp(x^2+4*
x)+x-exp(3))^3,x, algorithm="fricas")

[Out]

((x^2 + 1)*log(x - e^3 + e^(x^2 + 4*x) - log(x))^2 - 25*x)/(x*log(x - e^3 + e^(x^2 + 4*x) - log(x))^2)

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giac [B]  time = 3.53, size = 75, normalized size = 2.68 \begin {gather*} \frac {x^{2} \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2} + \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2} - 25 \, x}{x \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-1)*log(x)+(-x^2+1)*exp(x^2+4*x)+(x^2-1)*exp(3)-x^3+x)*log(-log(x)+exp(x^2+4*x)+x-exp(3))^3+(-
100*x^3-200*x^2)*exp(x^2+4*x)-50*x^2+50*x)/(x^2*log(x)-x^2*exp(x^2+4*x)+x^2*exp(3)-x^3)/log(-log(x)+exp(x^2+4*
x)+x-exp(3))^3,x, algorithm="giac")

[Out]

(x^2*log(x - e^3 + e^(x^2 + 4*x) - log(x))^2 + log(x - e^3 + e^(x^2 + 4*x) - log(x))^2 - 25*x)/(x*log(x - e^3
+ e^(x^2 + 4*x) - log(x))^2)

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maple [A]  time = 0.05, size = 32, normalized size = 1.14

method	result	size
risch	\(\frac {x^{2}+1}{x}-\frac {25}{\ln \left (x +{\mathrm e}^{\left (4+x \right ) x}-{\mathrm e}^{3}-\ln \relax (x )\right )^{2}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x^2-1)*ln(x)+(-x^2+1)*exp(x^2+4*x)+(x^2-1)*exp(3)-x^3+x)*ln(-ln(x)+exp(x^2+4*x)+x-exp(3))^3+(-100*x^3-2
00*x^2)*exp(x^2+4*x)-50*x^2+50*x)/(x^2*ln(x)-x^2*exp(x^2+4*x)+x^2*exp(3)-x^3)/ln(-ln(x)+exp(x^2+4*x)+x-exp(3))
^3,x,method=_RETURNVERBOSE)

[Out]

(x^2+1)/x-25/ln(x+exp((4+x)*x)-exp(3)-ln(x))^2

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maxima [B]  time = 0.53, size = 56, normalized size = 2.00 \begin {gather*} \frac {{\left (x^{2} + 1\right )} \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2} - 25 \, x}{x \log \left (x - e^{3} + e^{\left (x^{2} + 4 \, x\right )} - \log \relax (x)\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x^2-1)*log(x)+(-x^2+1)*exp(x^2+4*x)+(x^2-1)*exp(3)-x^3+x)*log(-log(x)+exp(x^2+4*x)+x-exp(3))^3+(-
100*x^3-200*x^2)*exp(x^2+4*x)-50*x^2+50*x)/(x^2*log(x)-x^2*exp(x^2+4*x)+x^2*exp(3)-x^3)/log(-log(x)+exp(x^2+4*
x)+x-exp(3))^3,x, algorithm="maxima")

[Out]

((x^2 + 1)*log(x - e^3 + e^(x^2 + 4*x) - log(x))^2 - 25*x)/(x*log(x - e^3 + e^(x^2 + 4*x) - log(x))^2)

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mupad [B]  time = 1.10, size = 29, normalized size = 1.04 \begin {gather*} x-\frac {25}{{\ln \left (x-{\mathrm {e}}^3-\ln \relax (x)+{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{x^2}\right )}^2}+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x + log(x - exp(3) + exp(4*x + x^2) - log(x))^3*(x - exp(4*x + x^2)*(x^2 - 1) + log(x)*(x^2 - 1) - x^3
+ exp(3)*(x^2 - 1)) - exp(4*x + x^2)*(200*x^2 + 100*x^3) - 50*x^2)/(log(x - exp(3) + exp(4*x + x^2) - log(x))
^3*(x^2*log(x) + x^2*exp(3) - x^2*exp(4*x + x^2) - x^3)),x)

[Out]

x - 25/log(x - exp(3) - log(x) + exp(4*x)*exp(x^2))^2 + 1/x

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sympy [A]  time = 2.57, size = 26, normalized size = 0.93 \begin {gather*} x - \frac {25}{\log {\left (x + e^{x^{2} + 4 x} - \log {\relax (x )} - e^{3} \right )}^{2}} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-1)*ln(x)+(-x**2+1)*exp(x**2+4*x)+(x**2-1)*exp(3)-x**3+x)*ln(-ln(x)+exp(x**2+4*x)+x-exp(3))**
3+(-100*x**3-200*x**2)*exp(x**2+4*x)-50*x**2+50*x)/(x**2*ln(x)-x**2*exp(x**2+4*x)+x**2*exp(3)-x**3)/ln(-ln(x)+
exp(x**2+4*x)+x-exp(3))**3,x)

[Out]

x - 25/log(x + exp(x**2 + 4*x) - log(x) - exp(3))**2 + 1/x

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